Question

Hi,

I am doing some revision on Fourier Transform on my own and I got stuck on this question for five days now. Wonder if you can help me.
Engineering Mathematics 4E, Anthony Croft
Ex, 24.8 Q1

phs overlap for S o. Hence 24.9 The discrete Fourier transform 745 EXERCISES 24.8 Find the convolution of Prove that convolution is commutative, tat is f*g=g*f,Note that correlation is not. Show that if f(r) and g(t) are both zero, for 0. and 6 Prove the convolution theorem. s)0 otherwise convolution of a function with itself is known as 71 Show that f(t) * g(r) = f(r) * g(-t). Deduce that a correlation can be expressed in terms of a convolution. Show also that f(t) * g(f) = f(t) * g(-t). volution. Find the autoconvolution f *f when 8 Prove that the correlation integral f gta-t)då can also be written in the form fooo g(λ) f(t + λ) da. f(t) otherwise 0 ion for the correlation is when t 0 hen 0 ansforms of f and g are Find the correlation of fu) I for-1t 1 and Prove the correlation theorem. zero otherwise, and gt)u()e-. Verify the correlation theorem for these functions Solutions 2 +jo 0 otherwise (e-e -1 jo) nsform of f 12-33IS6 t>1 g, obtained in part (a), we have 1> 6

I want to understand how to set the lower and upper bound in each "range" by applying the floating "t" value.
And how does sliding the triangle or the square block affect the convolution process.
Does fourier transform gives us greater flexibility than laplace transform for allowing the whole domain from -infinity to +infinity to be considered?

Many many thanks and much appreciated for your time and patience.

Kai

Answer 1

Given: f(t) = { 2/3 t: 0 lessthanorequalto t lessthanorequalto 0: otherwise g(t) = { 4: - 1 lessthanorequalto t lessthanorequalto 3 0: otherwise f(t) times g(T) = integral^infinity_- infinity f(t) g(t - tau) d tau limits are: - 0 - 1 = -1 to 3 - 1 = 2 3 - 2 = 2 to 0 + 3 = 3 0 + 3 = 3 + 0 3 + 3 = 6 justified i the when 1 + t < 0 implies t < - 1 The product g(t - tau) . t(tau) = 0 Therefore y(t) = g(t) times f(t) = f(t) times g(t) = 0 for t < - 1 case (ii): 1 + t < 3 implies t < 2 implies - 1 lessthanorequalto t lessthanorequalto 2 y(t) = integral^t + 1_0 g(t - tau) f (Tau) d tau = integral^t + 1_a 4 times 2/3 tau dt = 4/3 t^2]^t + 1_0 y (t) = 4/3 (t + 1)^2 Therefore y (t) = 4/3 (t + 1)^2 for - 1 lessthanorequalto t lessthanorequalto 2 cse (iii): - 2 < t lessthanorequalto 3 form the graph y(t) = integral^3_0 2/3 tau times 4 d tau = 4/3 tau^2]^3_0 y(T) = 12 Therefor